Digital number system, Binary Addition, Complemented number system, 1’s Complement, 2’s Complement, 8421 BCD Code, Gray Code, Excess-3 Code , Boolean Algebra, DeMorgan's Theorem

1. Digit’s 01010

2. NUMBER SYSTEM
BASE
Symbol
Intro…

3. Common Number Systems
System Base Symbols
Used by
humans?
Used in
computers?
Decimal 10 0, 1, … 9 Yes No
Binary 2 0, 1 No Yes
Octal 8 0, 1, … 7 No No
Hexa-decimal
16 0, 1, … 9,
A, B, … F
No No

4. Conversion Among Bases
 The possibilities:
Decimal Octal
Hexadecimal
Binary

5. Quick Example
2510 = 110012 = 318 = 1916
Base/Radix

6. Decimal to Decimal (just for fun)
Decimal Octal
Hexadecimal
Binary

7. 12510 => 5 x 100= 5
2 x 101= 20
1 x 102= 100
125
Base
Weight

8. Binary to Decimal
Hexadecimal
Decimal
Octal
Binary

9. Binary to Decimal
 Technique
 Multiply each bit by 2n, where n is the “weight” of the bit
 The weight is the position of the bit, starting from 0 on the right
 Add the results

10. Example
1010112 => 1 x 20 = 1
1 x 21 = 2
0 x 22 = 0
1 x 23 = 8
0 x 24 = 0
1 x 25 = 32
4310
Bit “0”

11. Octal to Decimal
Decimal Octal
Hexadecimal
Binary

12. Octal to Decimal
 Technique
 Multiply each bit by 8n, where n is the “weight” of the bit
 The weight is the position of the bit, starting from 0 on the right
 Add the results

13. Example
7248 => 4 x 80 = 4
2 x 81 = 16
7 x 82 = 448
46810

14. Hexadecimal to Decimal
Decimal Octal
Hexadecimal
Binary

15. Hexadecimal to Decimal
 Technique
 Multiply each bit by 16n, where n is the “weight” of the bit
 The weight is the position of the bit, starting from 0 on the right
 Add the results

16. Example
ABC16 => C x 160 = 12 x 1 = 12
B x 161 = 11 x 16 = 176
A x 162 = 10 x 256 = 2560
274810

17. Decimal to Binary
Decimal Octal
Hexadecimal
Binary

18. Decimal to Binary
 Technique
 Divide by two, keep track of the remainder
 First remainder is bit 0 (LSB, least-significant bit)
 Second remainder is bit 1
 Etc.

19. Example
12510 = ?2
2 125
2 6 2 1
2 3 1 0
2 1 5 1
2 7 1
2 3 1
2 1 1
0 1
12510 = 11111012

20. Octal to Binary
Decimal Octal
Hexadecimal
Binary

21. Octal to Binary
 Technique
 Convert each octal digit to a 3-bit equivalent binary representation

22. Example
7058 = ?2
7 0 5
111 000 101
7058 = 1110001012

23. Hexadecimal to Binary
Decimal Octal
Hexadecimal
Binary

24. Hexadecimal to Binary
 Technique
 Convert each hexadecimal digit to a 4-bit equivalent binary
representation

25. Example
10AF16 = ?2
1 0 A F
0001 0000 1010 1111
10AF16 = 00010000101011112

26. Decimal to Octal
Decimal Octal
Hexadecimal
Binary

27. Decimal to Octal
 Technique
 Divide by 8
 Keep track of the remainder

28. Example
123410 = ?8
8 1234
8 154 2
19 2
8
2 3
8
0 2
123410 = 23228

29. Decimal to Hexadecimal
Decimal Octal
Hexadecimal
Binary

30. Decimal to Hexadecimal
 Technique
 Divide by 16
 Keep track of the remainder

31. Example
123410 = ?16
123410 = 4D216
16 1234
77 2
16
4 13 = D
16
0 4

32. Binary to Octal
Decimal Octal
Hexadecimal
Binary

33. Binary to Octal
 Technique
 Group bits in threes, starting on right
 Convert to octal digits

34. Example
10110101112 = ?8
1 011 010 111
1 3 2 7
10110101112 = 13278

35. Binary to Hexadecimal
Decimal Octal
Hexadecimal
Binary

36. Binary to Hexadecimal
 Technique
 Group bits in fours, starting on right
 Convert to hexadecimal digits

37. Example
10101110112 = ?16
10 1011 1011
2 B B
10101110112 = 2BB16

38. Octal to Hexadecimal
Decimal Octal
Hexadecimal
Binary

39. Octal to Hexadecimal
 Technique
 Use binary as an intermediary

40. Example
10768 = ?16
1 0 7 6
001 000 111 110
2 3 E
10768 = 23E16

41. Hexadecimal to Octal
Decimal Octal
Hexadecimal
Binary

42. Hexadecimal to Octal
 Technique
 Use binary as an intermediary

43. Example
1F0C16 = ?8
1 F 0 C
0001 1111 0000 1100
1F0C16 = 174148
1 7 4 1 4

44. Exercise – Convert ...
Decimal Binary Octal
Don’t use a calculator!
Hexa-decimal
33
1110101
703
1AF
Skip answer Answer

45. Exercise – Convert …
Answer
Decimal Binary Octal
Hexa-decimal
33 100001 41 21
117 1110101 165 75
451 111000011 703 1C3
431 110101111 657 1AF

46. Common Powers (1 of 2)
 Base 10
Power Preface Symbol
10-12 pico p
10-9 nano n
10-6 micro 
10-3 milli m
103 kilo k
106 mega M
109 giga G
1012 tera T
Value
.000000000001
.000000001
.000001
.001
1000
1000000
1000000000
1000000000000

47. Common Powers (2 of 2)
 Base 2
Power Preface Symbol
210 kilo k
220 mega M
230 Giga G
Value
1024
1048576
1073741824
• What is the value of “k”, “M”, and “G”?
• In computing, particularly w.r.t. memory,
the base-2 interpretation generally applies

48. Review – multiplying powers
 For common bases, add powers
ab  ac = ab+c
26  210 = 216 = 65,536
or…
26  210 = 64  210 = 64k

49. Binary Addition (1 of 2)
 Two 1-bit values
A B A + B
0 0 0
0 1 1
1 0 1
1 1 10
“two”

50. Binary Addition (2 of 2)
 Two n-bit values
 Add individual bits
 Propagate carries
 E.g.,
10101 21
+ 11001 + 25
101110 46

51. Multiplication (1 of 3)
 Decimal (just for fun)
35
x 105
175
000
35
3675

52. Multiplication (2 of 3)
 Binary, two 1-bit values
A B A  B
0 0 0
0 1 0
1 0 0
1 1 1

53. Multiplication (3 of 3)
 Binary, two n-bit values
 As with decimal values
 E.g.,
1110
x 1011
1110
1110
0000
1110
10011010

54. Fractions
 Decimal to decimal (just for fun)
3.14 => 4 x 10-2 = 0.04
1 x 10-1 = 0.1
3 x 100 = 3
3.14

55. Fractions
 Binary to decimal
10.1011 => 1 x 2-4 = 0.0625
1 x 2-3 = 0.125
0 x 2-2 = 0.0
1 x 2-1 = 0.5
0 x 20 = 0.0
1 x 21 = 2.0
2.6875

56. Fractions
 Decimal to binary
3.14579
.14579
x 2
0.29158
x 2
0.58316
x 2
1.16632
x 2
0.33264
x 2
0.66528
x 2
1.33056
etc.
11.001001...

57. Exercise – Convert ...
Decimal Binary Octal
Don’t use a calculator!
Hexa-decimal
29.8
101.1101
3.07
C.82

58. Exercise – Convert …
Answer
Decimal Binary Octal
Hexa-decimal
29.8 11101.110011… 35.63… 1D.CC…
5.8125 101.1101 5.64 5.D
3.109375 11.000111 3.07 3.1C
12.5078125 1100.10000010 14.404 C.82

59. Complemented number system

60. Sign-Magnitude Notation
The simplest form of representation that employs a sign bit, the
leftmost significant bit. For an N-bit word, the rightmost N-1
bits hold the magnitude of the integer. Thus,
 00010010 = +18
 10010010 = -18

61. Drawbacks of Sign-Magnitude
Notation
Addition and subtraction require a consideration of
both the signs of the numbers, and their relative
magnitudes, in order to carry out the requested
operation.

62. 1’s Complement Notation
 Positive integers are represented in the same way as sign-magnitude
notation.
 A negative integer is represented by the 1's complement of
the positive integer in sign-magnitude notation.
 The ones complement of a number is obtained by
complementing each one of the bits, i.e., a 1 is replaced by a
0, and a 0 is replaced by a 1.
 18 (Base 10) = 00010010
 -18 = 1's complement of 18 = 11101101

63. 2’s Complement Notation
The 2's complement representation of positive
integers is the same as in sign-magnitude
representation.
A negative number is represented by the 2's
complement of the positive integer with the same
magnitude.
1. Perform the 1's complement operation.
2. Treating the result as an unsigned binary integer,
add 1.

64. 2’s Complement Example
18 = 00010010
1's complement = 11101101
+1
________
2’S of 18 = 11101110 = -18d

65. Binary Addition and Subtration
The simplest implementation is one in which the
numbers involved can be treated as unsigned
integers for purposes of addition.
1’s complement Addition
2’s complement Addition

66. 1’s complement Addition
 Consider the Subtrahend & Minuend.
 Calculate the 1’s complement of the Subtrahend number.
 Add it to the Minuend
 If the resulting is producing any carry then add the carry to
the LSB of the Result the sign is same as of the Minuend.
 If there is no carry then again take the 1’s complement of the
result and place minus sign before the result.
 This is the Require Result.

67. Example
ex.
+1 0001 Minuend
-6 0110 Subtrahend
1’s complement of 6 = 1001
Add : 0001 Minuend
1001 1’s Complement of Subtrahend
Result =1010 (No carry)
Again take the 1’ s Complement of 1010 = - 0101
Answer = > -5

68. Ones Complement End Around Carry
Perform the operation
-2 + -4
1’s complement of -2 = 1101
1’s complement of -4 = 1011
-------
1)1000
--> 1
--------
1001 = -6

69. 2’s complement Addition
 Calculate the 2’s complement of the Subtrahend number.
 Add the 2’s complement to the Minuend Number
 If the resultant has a carry discard it & this is the final Ans.
 Else if there is no carry calculate the 2’s complement of the
Resultant.
 This is the final Answer.

70. 2’s Complement Addition
: a)5 - 7
Binary equivalent of 7 = 0111
1’s complement of 7 = 1000
2’s complement of 7 = 1001
Now add 5 = 0101
+ (-)7 = 1001
1110 = -2
 In 2’s Complement addition, the Carry Out of the most significant bit is
ignored!

71. 8421 BCD Code
 In the 8421 Binary Coded Decimal (BCD) representation each
decimal digit is converted to its 4-bit pure binary equivalent.
 This coding is an example of a binary coded (each decimal
number maps to four bits) weighted (each bit represents a
number: 1, 2, 4, etc.) code.
 57dec = 0101 0111bcd

72. Questions
 Q. What is the decimal number for the following 8421 BCD
Code 0001 1001 0111 0010bcd?
 A. 1972
 Q. What is the 8421 BCD Code for the decimal number 421?
 A. 0100 0010 0001

73. Gray Code
 Gray coding is used for its speed & freedom from errors.
 In BCD or 8421 BCD when counting from 7 (0111) to 8 (1000)
requires 4 bits to be changed simultaneously.
 If this does not happen then various numbers could be
momentarily generated during the transition so creating
spurious numbers which could be read.
 Gray coding avoids this since only one bit changes between
subsequent numbers. Two simple rules.
 1. Start with all 0s.
 2. Proceed by changing the least significant bit (lsb) which will bring
about a new state.

74. Gray Code Continued
Decimal Gray Code
0 0000
1 0001
2 0011
3 0010
4 0110
5 0111
6 0101
7 0100
8 1100
9 1101
10 1111
11 1110
12 1010
13 1011
14 1001
15 1000

75. Excess-3 Code
 Excess-3 is a non weighted code used to express decimal
numbers. The code derives its name from the fact that each
binary code is the corresponding 8421 code plus 0011(3).
 1000 of 8421 = 1011 in Excess-3

76. ASCII Code
 ASCII: American Standard Code for Information Interchange
 The standard ASCII code defines 128 character codes (from 0
to 127), of which, the first 32 are control codes (non-printable),
and the other 96 are representable characters.
 In addition to the 128 standard ASCII codes there are other
128 that are known as extended ASCII, and that are platform-dependent.

77. ASCII Code Table

78. Boolean Algebra

79. Boolean function
• Boolean function: Mapping from Boolean variables to
a Boolean value.
• Truth table:
Represents relationship between a Boolean function and
its binary variables.
It enumerates all possible combinations of arguments and
the corresponding function values.

80. Boolean function and logic diagram
• Boolean algebra: Deals with binary variables and logic
operations operating on those variables.
• Logic diagram: Composed of graphic symbols for logic gates. A
simple circuit sketch that represents inputs and outputs of
Boolean functions.

81. Gates
 Refer to the hardware to implement Boolean operators.
 The most basic gates are

82. Boolean function and truth table

83. Basic identities of boolean algebra
• A Boolean algebra is a closed algebraic system containing set
of elements and the operators.
• The . (dot) operator is known as logical And.
• The + (plus) which refer to logical OR.

84. Basic Identities of Boolean Algebra
(1) x + 0 = x
(2) x · 0 = 0
(3) x + 1 = 1
(4) x · 1 = x
(5) x + x = x
(6) x · x = x
(7) x + x’ = 1
(8) x · x’ = 0
(9) x + y = y + x
(10) xy = yx
(11) x + ( y + z ) = ( x + y ) + z
(12) x (yz) = (xy) z
X I/P O/P = X + I/P
0 0 0 = > X
1 0 1 => X
X I/P O/P = X . I/P
0 0 0
1 0 0
X I/P O/P = X . I/P
0 1 0 => X
1 1 1 => X

85. Basic Identities of Boolean Algebra (DeMorgan’s
Theorem)
(15) ( x + y )’ = x’ y’
(16) ( xy )’ = x’ + y’
(17) (x’)’ = x

86. Function Minimization using Boolean Algebra
 Examples:
(a) a + ab = a(1+b)=a
(b) a(a + b) = a.a +ab=a+ab=a(1+b)=a.
(c) a + a'b = (a + a')(a + b)=1(a + b) =a+b
(d) a(a' + b) = a. a' +ab=0+ab=ab

87. Try
 F = abc + abc’ + a’c

88. The other type of question
Show that;
1- ab + ab' = a
2- (a + b)(a + b') = a 1- ab + ab' = a(b+b') = a.1=a
2- (a + b)(a + b') = a.a +a.b' +a.b+b.b'
= a + a.b' +a.b + 0
= a + a.(b' +b) + 0
= a + a.1 + 0
= a + a = a

89. More Examples
 Show that;
(a) ab + ab'c = ab + ac
(b) (a + b)(a + b' + c) = a + bc
(a) ab + ab'c = a(b + b'c)
= a((b+b').(b+c))=a(b+c)=ab+ac
(b) (a + b)(a + b' + c)
= (a.a + a.b' + a.c + ab +b.b' +bc)
= (a + ab’ + a.c + a.b + 0 + b.c)
=( a(1+b’ + c + b)+ b.c)
= ( a+ b.c)

90. Principle of Duality
 The dual of a statement S is obtained by interchanging . and
+; 0 and 1.
 Dual of (a*1)*(0+a’) = 0 is (a+0)+(1*a’) = 1
 Dual of any theorem in a Boolean Algebra is also a theorem.
 This is called the Principle of Duality.

91. DeMorgan's Theorem
(a) (a + b)' = a'b'
(b) (ab)' = a' + b'
Generalized DeMorgan's Theorem
(a) (a + b + … z)' = a'b' … z'
(b) (a.b … z)' = a' + b' + … z‘

92. DeMorgan's Theorem
 F = ab + c’d’
 F’ = ??
 F = ab + c’d’ + b’d
 F’ = ??
 Show that: (a + b.c)' = a'.b' + a'.c'

93. More DeMorgan's example
Show that: (a(b + z(x + a')))' =a' + b' (z' + x')
Sol: (a(b + z(x + a')))' = a' + (b + z(x + a'))'
= a' + b' (z(x + a'))'
= a' + b' (z' + (x + a')')
= a' + b' (z' + x'(a')')
= a' + b' (z' + x'a)
=a‘+b' z' + b'x'a
=(a‘+ b'x'a) + b' z'
=(a‘+ b'x‘)(a +a‘) + b' z'
= a‘+ b'x‘+ b' z‘
= a' + b' (z' + x')

94. More Examples
(a(b + c) + a'b)'=b'(a' + c')
ab + a'c + bc = ab + a'c
(a + b)(a' + c)(b + c) = (a + b)(a' + c)